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c)
\(4\left(3x-4\right)-2=18\)
<=> \(12x-16-2=18\)
<=> \(12x=36\)
<=> \(x=3\)
Vậy x=3
d)
\(\left(3x-10\right):10=50\)
<=> \(3x-10=500\)
<=> \(3x=510\)
<=> x= \(170\)
Vậy x= 170
f)
\(x-\left[42+\left(-25\right)\right]=-8\)
<=> \(x-17=-8\)
<=> x= \(9\)
Vậy x=9
h)
\(x+5=20-\left(12-7\right)\)
<=> \(x+5=15\)
<=> \(x=10\)
Vậy x= 10
k)
\(\left|x-5\right|=7-\left(-3\right)\)
<=> \(\left|x-5\right|=10\)
* Với \(x>=5\) ; ta được:
\(x-5=10\)
<=> x= 15 (thoả mãn điều kiện )
*Với \(x< 5\) ; ta được:
\(-\left(x-5\right)=10\)
<=> \(-x+5=10\)
<=> \(-x=5\)
<=> \(x=-5\) (thoả mãn điều kiện)
Vậy x=15 ; x= -5
i)
\(\left|x-5\right|=\left|7\right|\)
<=> \(\left|x-5\right|=7\)
*Với \(x>=5\) ; ta được:
\(x-5=7\)
<=> \(x=12\) (thoả mãn)
*Với \(x< 5\) ; ta được:
\(-\left(x-5\right)=7\)
<=> \(-x=2\)
<=> \(x=-2\) (thoả mãn)
Vậy x= 12; x= -2
m)
\(2^{x+1}.2^{2009}=2^{2010}\)
<=> \(2^{x+1+2009}=2^{2010}\)
<=> \(2^{x+2010}=2^{2010}\)
=> \(x+2010=2010\)
=> \(x=0\)
Vậy x=0
n)
\(10-2x=25-3x\)
<=>\(x=15\)
Vậy x=15
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75 * (3x-23) = 74 * 72
3x -23 = 74 * 72 : 75
3x -23 = 7
3x = 7 - 8
3x = -1
x = \(\frac{-1}{3}\)
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a: \(\dfrac{2}{3}x-\dfrac{1}{2}=\dfrac{1}{10}\)
nên \(\dfrac{2}{3}x=\dfrac{1}{10}+\dfrac{1}{2}=\dfrac{6}{10}=\dfrac{3}{5}\)
hay \(x=\dfrac{3}{5}:\dfrac{2}{3}=\dfrac{9}{10}\)
b: \(\Leftrightarrow5-\dfrac{4}{7}x=13\)
nên 4/7x=-8
hay x=-12
c: \(\left(x+\dfrac{1}{2}\right)\cdot\left(\dfrac{2}{3}-2x\right)=0\)
=>x+1/2=0 hoặc 2/3-2x=0
=>x=-1/2 hoặc x=1/3
d: \(\dfrac{2}{3}x-\dfrac{1}{2}x=\dfrac{5}{12}\)
nên 1/6x=5/12
hay x=5/2
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\dfrac{2}{3}x-\dfrac{1}{2}=\dfrac{1}{10}\)
\(\dfrac{2}{3}x=\dfrac{1}{10}+\dfrac{1}{2}=\dfrac{3}{5}\)
\(x=\dfrac{3}{5}:\dfrac{2}{3}=\dfrac{9}{10}\)
b) \(\dfrac{39}{7}:x=13\)
\(x=\dfrac{\dfrac{39}{7}}{13}=\dfrac{3}{7}\)
c) \(\left(\dfrac{14}{5}x-50\right):\dfrac{2}{3}=51\)
\(\dfrac{14}{5}x-50=51\cdot\dfrac{2}{3}=34\)
\(\dfrac{14}{5}x=34+50=84\)
\(x=\dfrac{84}{\dfrac{14}{5}}=30\)
d) \(\left(x+\dfrac{1}{2}\right)\left(\dfrac{2}{3}-2x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=0\\\dfrac{2}{3}-2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\)
e) \(\dfrac{2}{3}x-\dfrac{1}{2}x=\dfrac{5}{12}\)
\(\dfrac{1}{6}x=\dfrac{5}{12}\)
\(x=\dfrac{5}{12}:\dfrac{1}{6}=\dfrac{5}{2}\)
g) \(\left(x\cdot\dfrac{44}{7}+\dfrac{3}{7}\right)\dfrac{11}{5}-\dfrac{3}{7}=-2\)
\(\left(x\cdot\dfrac{44}{7}+\dfrac{3}{7}\right)\cdot\dfrac{11}{5}=-2+\dfrac{3}{7}=-\dfrac{11}{7}\)
\(x\cdot\dfrac{44}{7}+\dfrac{3}{7}=-\dfrac{11}{7}:\dfrac{11}{5}=-\dfrac{5}{7}\)
\(\dfrac{44}{7}x=-\dfrac{5}{7}-\dfrac{3}{7}=-\dfrac{8}{7}\)
\(x=-\dfrac{8}{7}:\dfrac{44}{7}=-\dfrac{2}{11}\)
h) \(\dfrac{13}{4}x+\left(-\dfrac{7}{6}\right)x-\dfrac{5}{3}=\dfrac{5}{12}\)
\(\dfrac{25}{12}x-\dfrac{5}{3}=\dfrac{5}{12}\)
\(\dfrac{25}{12}x=\dfrac{5}{12}+\dfrac{5}{3}=\dfrac{25}{12}\)
\(x=1\)
Mỏi tay woa bn làm nốt nha!!
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1, 4[3x + 4] = 12x + 16
2, 4/3[5/3x + 6] = 20/9x + 12
3, [4/3x + 7]. 5 + 2 = 20/3x + 35 + 2 = 20/3x + 37
4, [3/7x - 6][-7] + 3x = -3x + 42 + 3x = 42
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3x - 7 = 2x + 5
3x - 2x = 5 + 7
x = 12
|3x - 2| = 7
\(\Rightarrow\left\{\begin{matrix}3x-2=7\\-\left(3x-2\right)=7\end{matrix}\right.\Rightarrow\left\{\begin{matrix}3x=9\\-3x+2=7\end{matrix}\right.\)
\(\Rightarrow\left\{\begin{matrix}x=3\\-3x=5\end{matrix}\right.\Rightarrow\left\{\begin{matrix}x=3\\x=-\frac{5}{3}\end{matrix}\right.\)
4 - |x - 2| = -3
|x - 2| = 4 - (-3)
|x - 2| = 7
\(\Rightarrow\left\{\begin{matrix}x-2=7\\-\left(x-2\right)=7\end{matrix}\right.\Rightarrow\left\{\begin{matrix}x=9\\-x+2=7\end{matrix}\right.\)
\(\Rightarrow\left\{\begin{matrix}x=9\\-x=5\end{matrix}\right.\Rightarrow\left\{\begin{matrix}x=9\\x=-5\end{matrix}\right.\)
|2x - 3| = x - 1
\(\Rightarrow\left\{\begin{matrix}2x-3=x-1\\-\left(2x-3\right)=x-1\end{matrix}\right.\Rightarrow\left\{\begin{matrix}2x-x=-1+3\\-2x+3=x-1\end{matrix}\right.\)
\(\Rightarrow\left\{\begin{matrix}x=2\\-2x-x=-1-3\end{matrix}\right.\Rightarrow\left\{\begin{matrix}x=2\\-3x=-4\end{matrix}\right.\)
\(\Rightarrow\left\{\begin{matrix}x=2\\x=\frac{4}{3}\end{matrix}\right.\)
|3x + 1| = x + 3
\(\Rightarrow\left\{\begin{matrix}3x+1=x+3\\-\left(3x+1\right)=x+3\end{matrix}\right.\Rightarrow\left\{\begin{matrix}3x-x=3-1\\-3x-1=x+3\end{matrix}\right.\)
\(\Rightarrow\left\{\begin{matrix}2x=2\\-3x-x=3+1\end{matrix}\right.\Rightarrow\left\{\begin{matrix}x=1\\-4x=4\end{matrix}\right.\)
\(\Rightarrow\left\{\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
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\(\left(3x-2^4\right)\cdot7=2\cdot7^2\)
\(\Leftrightarrow3x-2^4=2\cdot7^2:7\)
\(\Leftrightarrow3x-16=2\cdot7\)
\(\Leftrightarrow3x-16=14\)
\(\Leftrightarrow3x=14+16\)
\(\Leftrightarrow3x=30\)
\(\Leftrightarrow=30:3\)
\(\Leftrightarrow x=10\)
Vậy x=10